1) Red squares BN = AI + CE -- Pythagoras's theorem[br][br]2) Blue triangles AEH, CDN, BMI are all equal in area to ABC, reasoning via X and Y and base sides.[br][br]3) Green angles GHI and IBM are equal and green triangle GHI is congruent to IBM (side angle side), so IG = IK = IM. IH'K is congruent to IHK as angle HIK = angle HIG and the adjacent sides correspond. This means G and K are the same distance from the line HH', so GK is parallel to HI. Similarly, DE is parallel to PF and MN is parallel to LO.[br][br]4) GK = 4HI, because TU=HI, TG = AH (HTG congruent to EAH) and UK = UG (symmetry). Similarly, PF = 4DE. [br] Dark blue triangles IVK and LWM are equal, so WM = VK. Similarly, OX = QD (dark green triangles PQD and NXO are congruent). Also, WX=MJ and XN=NJ, so M and N are the midpoints of WJ and XJ. That makes WX=2MN, so LO = 4MN.[br] [br]5) Each of the trapezia we just looked at (HIKG, OLMN and PFED) have five times the area of ABC. [br][br]6) The areas of orange squares MK and NP are together five times EG. This is because: [br] the square on MI is (the square on MY) + (the square on IY) = (AC^2) + (2AB)^2 = 4AB^2 + AC^2. [br] the square on ND is (the square on NZ) + (the square on DZ) = (AB^2) + (2AC)^2 = 4AC^2 + AB^2[br] the sum of these is 5(AB^2 + AC^2) = 5BC^2, and BC = HE. [br] [br]7) A'S = A'T, so A'SAT is a square and the bisector of angle B'A'C' passes through A.[br] However, the bisectors of angle A'B'C' and A'C'B' do not pass through B and C (resp.) [Reasoning not immediately clear to me][br][br]8) Square LO = square GK + square FP, as LO = 4AC, GK = 4AB and FP = 4BC.[br][br]9) What the author means by 'etc. etc.', I don't know!